Ask question

Ask question

All categories

3 years ago

6

Calculate the impulse imparted when a 3,000-kg car hits a wall at 60 . m/s and comes to a stop.

1 answer:

Pani-rosa [81]3 years ago

7 0

Impulse = change in momentum

The answer is 0.

You might be interested in

Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity

Anna11 [10]

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$ which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

4 0

3 years ago

A bell is rung. What best describes the density of air around the bell? The air density does not change. The air density increas

dezoksy [38]

I think the closest possible answer to this question is The air density increases and decreases repeatedly before returning to normal.Thank you for your question. Please don't hesitate to ask in Brainly your queries.

7 0

3 years ago

Read 2 more answers

A roller coaster is at the top of a hill and rolls to the top of a lower hill if mechanical energy is constant then on the top o

mafiozo [28]

The kinetic energy is greater on the second hill

hope i helped have a great day

4 0

3 years ago

The initial speed of a body is 7.1 m/s. What is its speed after 2.23 s if it accelerates uniformly at 2.64 m/s 2 ? Answer in uni

Nana76 [90]

13.0m/s

1.2m/s

Explanation:

Given parameters:

Initial speed of the body = 7.1m/s

time taken = 2.23s

Acceleration = 2.64m/s²

Unknown:

Final speed = ?

Solution:

Acceleration is the rate of change of velocity with time.

a = $\frac{V - U}{T}$

a = acceleration

V = final speed

U = initial speed

T = time taken

Input the variables and solve for V;

2.64 = $\frac{V - 7.1}{2.23}$

V - 7.1 = 5.9 expression 1

V = 5.9 + 7.1 = 13.0m/s

B

Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;

from expression 1;

V - 7.1 = -5.9

V = -5.9 + 7.1 = 1.2m/s

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

5 0

3 years ago

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s

andre [41]

Answer:

$d=1.84\ mm$

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

$\displaystyle C=\frac{\epsilon_o A}{d}$

where

$\epsilon_o=8.85\cdot 10^{-12}\ F/m$

A = area of the plates = $\pi r^2$

d = separation of the plates

$\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}$

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

$\displaystyle X_c=\frac{1}{wC}$

where w is the angular frequency

$w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s$

Solving for C

$\displaystyle C=\frac{1}{wX_c}$

The reactance can be found knowing the total impedance of the circuit:

$Z^2=R^2+X_c^2$

Where R is the resistance, $R=15 K\Omega=15000\Omega$ . Solving for Xc

$X_c^2=Z^2-R^2$

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

$\displaystyle Z=\frac{V}{I}$

The rms current is the peak current Ip divided by $\sqrt{2}$ , thus

$\displaystyle Z=\frac{\sqrt{2}V}{I_p}$

$I_p=0.65\ mA/1000=0.00065\ A$

Now collect formulas

$\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2$

Or, equivalently

$\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}$

$\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}$

$X_c=36176.34\ \Omega$

The capacitance is now

$\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F$

The radius of the plates is

$r=18\ cm/2=9 \ cm = 0.09 \ m$

The separation between the plates is

$\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}$

$d=0.00184\ m$

$\boxed{d=1.84\ mm}$

8 0

3 years ago